• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

Prelim 2016 Maths Help Thread (4 Viewers)

Status
Not open for further replies.

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,805
Gender
Male
HSC
2015
Find general solution to sin(3x)+sin(x)=0
Use the fact that sin(3x)=3sin(x)-4sin3(x) to complete this question.

Proof is left as an exercise to the reader.....
 

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
I tried that method..it didnt work out for me
sin(3x) = sin(2x + x)

= sin(2x)cos(x) + cos(2x)sin(x).

Now, use double angle formulas (for the cos(2x), use 1 – 2sin2 x), and use Pythagorean identity when necessary.
 

Green Yoda

Hi Φ
Joined
Mar 28, 2015
Messages
2,859
Gender
Male
HSC
2017
Answer only says nπ
but I also got nπ+(-1)^n x (π/2) or nπ+(-1)^n x (-π/2) as the gen sol
 

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,805
Gender
Male
HSC
2015
Answer only says nπ
but I also got nπ+(-1)^n x (π/2) or nπ+(-1)^n x (-π/2) as the gen sol
They combine to give n.pi

Think about it.

(If you don't see it, start writing out the values for n=0,1,2,3,4,5,6,...)
 

Green Yoda

Hi Φ
Joined
Mar 28, 2015
Messages
2,859
Gender
Male
HSC
2017
They combine to give n.pi

Think about it.

(If you don't see it, start writing out the values for n=0,1,2,3,4,5,6,...)
ah I see, just curious..in exam conditions if you dont pick up this connection..do you loose marks for writing 3 separate?
 

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
Your answer would be n*pi/2 rather than n*pi, as you can see by splitting up into cases of n even (n = 2k), or n odd (n = 2k-1).
 

Trebla

Administrator
Administrator
Joined
Feb 16, 2005
Messages
8,383
Gender
Male
HSC
2006
Find general solution to sin(3x)+sin(x)=0
Another approach would be to note that this equation is equivalent to

sin 3x = sin(-x)

If you solve it within one 'revolution' (like you would for a standard trig equation) you get

3x = -x
or
3x = π - (-x)

Now generalise this to multiple 'revolutions'

3x = -x + 2πn
or
3x = π - (-x) + 2πn

This resolves to

x = πn/2
or
x = 3πn/2

However, x = 3πn/2 represents a subset of the values within x = πn/2 so the most general solution is x = πn/2
 

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,805
Gender
Male
HSC
2015
ah I see, just curious..in exam conditions if you dont pick up this connection..do you loose marks for writing 3 separate?
Doubt it.

And refer to what they^ said. I didn't actually check your answer nor the given ones.
 

fluffchuck

Active Member
Joined
Apr 29, 2016
Messages
250
Location
Sydney
Gender
Male
HSC
2017
Uni Grad
2021
Yep, pretty much what Trebla said,

sin(3x) + sin(x) = 0

sin(3x) = -sin(x)

And since the sine function is odd, sin(-x) = -sin(x)

Hence, sin(3x) = sin(-x)

You should be able to solve from there, using your general solution formula. :)
 

eyeseeyou

Well-Known Member
Joined
Nov 20, 2015
Messages
4,125
Location
Space
Gender
Undisclosed
HSC
N/A
More motion question for you all (this will be good for the current 3U kids):

This is motion defined by a=a(x)

1. The acceleration of particle P, moving in a straight line, is given by x (with 2 dots on top of x)=2x-3 where x metres is the displacement from the origin O. Initially the particle is at O and its velocity v is 2 metres per second
a. Show that the velocity of the particle v^2=2x^2-6x+4
b. Calculate the velocity and acceleration of P at x=1 and briefly describe the motion of P after it moves from x=1
2. The acceleration of a particle is (2x-5) m/s^2, where x in the distance in metres from the origin
a. Find an expression for the velocity of this particle in terms of x, given that the particle is at rest one metre to the left of the origin initially
b. Describe the motion
3. A particle is moving along a straight line, with velocity v m/s and acceleration given by the expression k(4-x^2) m/s^2 where k is a constant
a. Show that v^2=4+Ae^(-2kx) where A is a constant satisifies the acceleration condition
b. If it starts from x=0 with a velocity of 7m/s, find the value of A
c. Does the particle ever change direction? Justify your answer
d. At x=1, v=4m/s, find the speed correct to two decimal places when x=2
e. As the motion continues, what happens, to the velocity and acceleration
 

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
More motion question for you all (this will be good for the current 3U kids):

This is motion defined by a=a(x)

1. The acceleration of particle P, moving in a straight line, is given by x (with 2 dots on top of x)=2x-3 where x metres is the displacement from the origin O. Initially the particle is at O and its velocity v is 2 metres per second
a. Show that the velocity of the particle v^2=2x^2-6x+4
b. Calculate the velocity and acceleration of P at x=1 and briefly describe the motion of P after it moves from x=1
2. The acceleration of a particle is (2x-5) m/s^2, where x in the distance in metres from the origin
a. Find an expression for the velocity of this particle in terms of x, given that the particle is at rest one metre to the left of the origin initially
b. Describe the motion
3. A particle is moving along a straight line, with velocity v m/s and acceleration given by the expression k(4-x^2) m/s^2 where k is a constant
a. Show that v^2=4+Ae^(-2kx) where A is a constant satisifies the acceleration condition
b. If it starts from x=0 with a velocity of 7m/s, find the value of A
c. Does the particle ever change direction? Justify your answer
d. At x=1, v=4m/s, find the speed correct to two decimal places when x=2
e. As the motion continues, what happens, to the velocity and acceleration




 

Orwell

Well-Known Member
Joined
Dec 2, 2015
Messages
830
Gender
Male
HSC
2017
I'm so bad at visualising worded questions, especially bearings in trigonometry. Can anyone solve this and maybe send me a picture of the working out and the diagram they constructed?

 

Orwell

Well-Known Member
Joined
Dec 2, 2015
Messages
830
Gender
Male
HSC
2017
I found it in my Maths in Focus text book. Also, I'll probably be posting innumerable seemingly easy questions considering I have an exam tomorrow haha. I've got another question:
 

jathu123

Active Member
Joined
Apr 21, 2015
Messages
357
Location
Sydney
Gender
Male
HSC
2017
I found it in my Maths in Focus text book. Also, I'll probably be posting innumerable seemingly easy questions considering I have an exam tomorrow haha. I've got another question:
cos(θ+10) = -1/2 [10<=θ+10<=370]
acute angle is 60, therefore
θ+ 10 = (180-60), (180+60) [All Stations To Central]
θ + 10 = 120, 240
θ = 110, 230
 
Status
Not open for further replies.

Users Who Are Viewing This Thread (Users: 0, Guests: 4)

Top